In a poker game with three players A, B and C, the dealer is chosen by the following procedure. In the order A, B, C, A, B, C, . . ., a card from a well-shuffled deck is dealt to each player until someone gets an ace. This first player receiving an ace gets to start the game as dealer. Do you think that everyone has an equal chance to become dealer?
Let $A_i$ be the event that the first ace appairs at $i$th attempt. Since $\mathbb{P}(A_k)=\frac{48}{52}\cdot \frac{47}{51}\cdot ... \cdot \frac{48-k+2}{52-k+2}\cdot \frac{4}{52-k+1}$ decreases for $k\rightarrow \infty$ and A starts the game before the others, the probability that A becomes a dealer would be greater than the probability for B and in turn the probability for B would be greater than the probability for C.
So defining $A=[$A becomes a dealer$]$ and $A_i=[$A become dealer at $i$th turn$]$, can I use $\mathbb{P}(A)=\mathbb{P}(\bigcup_{i=0}^{\infty}(A\cap A_i))=\sum_{i=0}^{\infty}\mathbb{P}(A_i)\mathbb{P}(A)$ to calculate them? If not, how can I do?
Thanks in advance.
Here is how the probability will look for each turn -
Probability of A winning in 1st turn $P_A(1) = \dfrac{4}{52}$
Probability of nobody winning in 1st turn $Q(1) = \dfrac{48 \times 47 \times 46}{52 \times 51 \times 50} = \dfrac{48! \times 49!}{52! \times 45!}$
Probability of A winning in 2nd turn $P_A(2) = Q(1) \times \dfrac{4}{49} = 4 \times \dfrac{48! \times 48!}{52! \times 45!}$
Similarly, $Q(2) = \dfrac{48! \times 46!}{52! \times 42!} \,$ and $\, P_A(3) = 4 \times \dfrac{48! \times 45!}{52! \times 42!}$
So, $P_A(k+1) = 4 \times \dfrac{48! \times (51-3k)!}{52! \times (48-3k)!}$ where $0 \le k \le 15$
When I put this in WolframAlpha, I get $\, \sum \limits_{k=0}^{15} P_A(k+1) \approx 0.36$. See below from WolframAlpha.